Maximizing an Integral | Putnam 2006 B5
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- Опубліковано 15 вер 2024
- In today's video I go over Putnam 2006 B5. This problem was about maximizing an integral, and a clever representation allowed it to be maximized. If you wanna see me solve some problem, let me know in the comments.
Related Links : / creative_math_
Putnam 2006 Paper : math.hawaii.edu...
#Calculus #Putnam #Integral
Euler-Lagrange equation also is a cheap way to arrive at the answer, tho this way is more clever and with Euler-Lagrange you have to prove the function is maximal somehow anyways. Subbed!
Oh you're right it directly gives us f(x) = x/2
I think if you take f(x) = 1 you get - 1/6 as the result of the integral so you can be sure it's not a minimum, and since it's an extremal maybe you can then say it can only be a maximum 🤷♂️
Nice method
You could also define another function h(y)= integral from 0 to 1
(x^2y-xy^2)dx
Then since we need Max we can equate h'(y)=0
So integral 0 to 1 ( x^2-2xy)=0
Therefore y=x/2
Thanks upload more videos 😎🔥🔥😊
I like it
No shit